Answer
$$
\lim _{x \rightarrow \infty} \left(\frac{x+a}{x-a}\right)^{x}=e
$$
This equation is true when $ a=\frac{1}{2}$.
Work Step by Step
$$
\lim _{x \rightarrow \infty} \left(\frac{x+a}{x-a}\right)^{x}=e
$$
Let
$$
L =\lim _{x \rightarrow \infty} \left(\frac{x+a}{x-a}\right)^{x}
$$
then
$$
\begin{aligned} \ln L &=\lim _{x \rightarrow \infty} \ln \left(\frac{x+a}{x-a}\right)^{x} \\
&=\lim _{x \rightarrow \infty} x \ln \left(\frac{x+a}{x-a}\right)\\
&=\lim _{x \rightarrow \infty} \frac{\ln (x+a)-\ln (x-a)}{1 / x} \\
& \quad\quad\quad\left[ \text { we can use L'Hospital's } \right] \\
&= \lim _{x \rightarrow \infty} \frac{\frac{1}{x+a}-\frac{1}{x-a}}{-1 / x^{2}} \\ &=\lim _{x \rightarrow \infty}\left[\frac{(x-a)-(x+a)}{(x+a)(x-a)} \cdot \frac{-x^{2}}{1}\right] \\
&=\lim _{x \rightarrow \infty} \frac{2 a x^{2}}{x^{2}-a^{2}}\\
&=\lim _{x \rightarrow \infty} \frac{2 a}{1-a^{2} / x^{2}} \\
&=2 a \end{aligned}
$$
Hence, $$
\ln L =2a, $$
so
$$
L =e^{2a}
$$
From the original equation, we want
$$
L =e^{1}
$$
$ \Rightarrow $
$$
2a=1 \Rightarrow a=\frac{1}{2}
$$
So, we obtain that the original equation is true when $ a=\frac{1}{2}$.
